Under the assumption of no arbitrage without vanish risk, in an incomplete market $(\Omega,{\cal F}, P)$, the set of equivalent martingale measure is NOT empty, i.e. ${\cal P} = \{Q: Q \sim P\}\neq \emptyset.$

My question is: in the following simplified market with one stock which is driving by two independent Brownian Motions and one bond, i.e.

How to calculate all the equivalent martingale ${\cal P}.$ We suppose that $\mu,\sigma_1,\sigma_2, r$ are constants.

One approach in my mind is using another stock to complete the market, i.e. we suppose there is another stock $\tilde{S}$ with parameters, $\tilde{\mu}, \tilde{\sigma_1},\tilde{ \sigma_2}$ such that
$$d\tilde{S_t} = \tilde{S_t}(\tilde{\mu} dt + \tilde{\sigma_1} dW_1(t) + \tilde{\sigma_2} dW_2(t)).$$

Then, following the classic method, we could get the equivalent martingale measures described by parameters, $\mu,\sigma_1,\sigma_2, r, \tilde{\mu}, \tilde{\sigma_1},\tilde{ \sigma_2}.$

But, how could I know the equivalent martingale measure obtained by above approach are the set of all the equivalent martingale measures in this financial market?

The question is ambiguously worded. If the "martingale" in "equivalent martingale measure" is understood relative to the filtration generated by the stock price process $S$, then there is but one equivalent martingale measure (EMM), as noted by @TheBridge. If the filtration $({\mathcal F}_t)_{t\ge 0}$ is taken to be that generated by $W_1$ and $W_2$, then there are many EMMs. Suppose (for simplicity) that $\mu=0$ and $\sigma_1=\sigma_2=1$. Then for each real $\alpha$ the measure $Q_\alpha$ defined on ${\mathcal F}_t$ by
$$
dQ_\alpha/dP := \exp(\alpha W_1(t)-\alpha W_2(t)- \alpha^2t)
$$
is an equivalent martingale measure. And there are more: If $H$ is a bounded $({\mathcal F}_t)$-predictable process then, on ${\mathcal F}_t$,
$$
dQ/dP :=\exp\left(\int_0^t H_s dW_1(s) - \int_0^t H_s dW_2(s) -\int_0^t H_s^2 ds\right)
$$
defines an EMM.